Fiber Product of Schemas I have a question about a equation in the proof of Prop. 3.1.16 in Liu's "Algebraic Geometry" (page 83):

$\DeclareMathOperator{\Spec}{Spec}$
Why the equation $(X \times _Y V) \times _V \Spec k(y) = f^{-1} (V) _y$ holds, where $Q \times _S R$ means the fiber product of $S$-schemes Q and R?
 A: $\DeclareMathOperator{\Spec}{Spec}$
I don't know what Liu had in mind, but you can argue directly that $X \times_Y \Spec \kappa(y)$ is isomorphic to $f^{-1}(V) \times_V \Spec \kappa(y)$ as follows.  (this is a bit technical and you might be better off just doing this yourself instead of going through what I wrote; I would draw a diagram to follow).
Let $f'$ be the morphism of schemes $f': f^{-1}(V) \rightarrow V$ coming from the morphism $f: X \rightarrow Y$.  Let $\iota: \Spec \kappa(y) \rightarrow Y$ be the structural morphism, and let $\iota'$ be the corresponding morphism $\iota': \Spec \kappa(y) \rightarrow V$.  Also, let $i: f^{-1}(V) \rightarrow X$ and $j: V \rightarrow Y$ be the inclusion morphisms.
Let $\pi_1, \pi_2$ be the projections on $X \times_Y \Spec \kappa(y)$, and $\pi_1', \pi_2'$ the projections on $f^{-1}(V) \times_V \Spec \kappa(y)$.
We have morphisms $i \circ \pi_1'$ and $\pi_2'$ from $f^{-1}(V) \times_V \Spec \kappa(y)$ to $X$ and $\Spec \kappa(y)$, respectively.  They are compatible with the structural morphisms $f: X \rightarrow Y$ and $\iota: \Spec \kappa(y) \rightarrow Y$, since
$$f \circ i \circ \pi_1' = j \circ f' \circ \pi_1' = j \circ \iota' \circ \pi_2' = \iota \circ \pi_2'$$
Therefore, by the universal property of $X \times_Y \Spec \kappa(y)$, there exists a unique morphism $\Phi: f^{-1}(V) \times_V \Spec \kappa(y) \rightarrow X \times_Y \Spec \kappa(y)$ such that $i \circ \pi_1' = \pi_1 \circ \Phi$ and $\pi_2' = \pi_2 \circ \Phi$.
I say $\Phi$ is an isomorphism of schemes.  By standard universal property arguments, this follows immediately from:

Claim: $f^{-1}(V) \times_V \Spec \kappa(y)$, together with the maps $i \circ \pi_1'$ and $\pi_2'$ to $X$ and $\Spec \kappa(y)$ respectively, is the fiber product of $X$ and $\Spec \kappa(y)$ over $Y$.

Proof of claim: Suppose $Z$ is a scheme, and $h_1: Z \rightarrow X, h_2: Z \rightarrow \Spec \kappa(y)$ are morphisms such that $f \circ h_1 = \iota \circ h_2$.  We need to show that there exists a unique morphism of schemes $h: Z \rightarrow f^{-1}(V) \times_V \Spec \kappa(y)$ such that $i \circ \pi_1' \circ h = h_1$ and $\pi_2' \circ h = h_2$.
The crucial observation is that, as maps of sets, since the image of $\iota$ is the single point $y$ of $Y$, the image of $h_1$ is contained in $f^{-1}\{y\} \subseteq f^{-1}(V)$.  A morphism of schemes whose set theoretic image is contained in open subset of a scheme is the same as a morphism of schemes into that open subscheme.  So there exists a unique morphism $\overline{h_1}: Z \rightarrow f^{-1}(V)$ such that $i \circ \overline{h_1} = h_1$.  Now,
$$j \circ f' \circ \overline{h_1} = f \circ i \circ \overline{h_1} = f \circ h_1 = \iota \circ h_2 = j \circ \iota' \circ h_2$$
so $$f' \circ \overline{h_1} = \iota' \circ h_2$$
as morphisms of schemes $Z \rightarrow f^{-1}(V)$.  So by the universal property of $f^{-1}(V) \times_V \Spec \kappa(y)$, there exists ($\ast$) a unique morphism $h: Z \rightarrow f^{-1}(V) \times_V \Spec \kappa(y)$ such that $\pi_1' \circ h = \overline{h_1}$ and $\pi_2' \circ h = h_2$.  Then
$$i \circ \pi_1' \circ h = i \circ \overline{h_1} = h \textrm{ and } \pi_2' \circ h = h_2 $$
which is what we wanted to show about $h$.  To see that $h$ is the unique morphism satisfying this property, suppose $H: Z \rightarrow f^{-1}(V) \times_V \Spec \kappa(y)$ is another morphism satisfying the same.  We can write $i \circ \pi_1' \circ H = i \circ \overline{h_1}$, and cancel the $i$ to get $\pi_1' \circ H = \overline{h_1}$ and $\pi_2' \circ H = h_2$.  By the uniqueness of $h$ in $\ast$, we get $H = h$.  $\blacksquare$
This reduces the Proposition to the case where $Y$ is affine: once you show that the underlying map of $\pi_1'$ is a homeomorphism onto the set of $x \in f^{-1}(V)$ such that $f(x) = y$, the fact that $\pi_1 \circ \Phi = \pi_1'$ as maps of sets into $X$ shows you that the same is true for $\pi_1$.
A: $\DeclareMathOperator{\Spec}{Spec}$Oh I get what he's saying now.  $X \times_Y V$ is isomorphic to $f^{-1}(V)$.  This implies that $(X \times_Y V) \times_V \Spec \kappa(y) \cong f^{-1}(V) \times_V \Spec \kappa(y) = f^{-1}(V)_y$. This can be seen in the same way as my previous answer, by showing that $f^{-1}(V)$, together with the maps $i:f^{-1}(V) \rightarrow X$ and $f': f^{-1}(V) \rightarrow V$, satisfies the required universal property.
If you want an intuitive reason for why  isomorphisms like this should hold, just imagine you're in the category of sets.  If $f: X \rightarrow Y$ is a map of sets, and $V \subseteq Y$, then the fiber product $X \times_Y V$ is literally 
$$X \times_Y V = \{ (x,v) \in X \times V : f(x) = v \}$$
which you can identify with $f^{-1}(V)$.
You can actually make this intuition precise by giving a different proof of these isomorphisms, using the Yoneda lemma.
